演算法 —— Count pairs with given sum
原文：https://www.geeksforgeeks.org/count-pairs-with-given-sum/
Given an array of integers, and a number ‘sum’, find the number of pairs of integers in the array whose sum is equal to ‘sum’.
Examples:
Input : arr[] = {1, 5, 7, -1},
sum = 6
Output : 2
Pairs with sum 6 are (1, 5) and (7, -1)
Input : arr[] = {1, 5, 7, -1, 5},
sum = 6
Output : 3
Pairs with sum 6 are (1, 5), (7, -1) &
(1, 5)
Input : arr[] = {1, 1, 1, 1},
sum = 2
Output : 6
There are 3! pairs with sum 2.
Input : arr[] = {10, 12, 10, 15, -1, 7, 6,
5, 4, 2, 1, 1, 1},
sum = 11
Output : 9Expected time complexity O(n)
Naive Solution – A simple solution is to traverse each element and check if there’s another number in the array which can be added to it to give sum.
// C# implementation of simple// method to find count of
// pairs with given sum.
using System;
class GFG {
public static void getPairsCount(int[] arr, int sum)
{
int count = 0; // Initialize result
// Consider all possible pairs
// and check their sums
for (int i = 0; i < arr.Length; i++)
for (int j = i + 1; j < arr.Length; j++)
if ((arr[i] + arr[j]) == sum)
count++;
Console.WriteLine("Count of pairs is " + count);
}
// Driver Code
static public void Main()
{
int[] arr = { 1, 5, 7, -1, 5 };
int sum = 6;
getPairsCount(arr, sum);
}
}
// This code is contributed
// by Sach_Code
|
Count of pairs is 3
Time Complexity: O(n2)
Auxiliary Space: O(1)
Efficient solution –
A better solution is possible in O(n) time. Below is the Algorithm –
- Create a map to store frequency of each number in the array. (Single traversal is required)
- In the next traversal, for every element check if it can be combined with any other element (other than itself!) to give the desired sum. Increment the counter accordingly.
- After completion of second traversal, we’d have twice the required value stored in counter because every pair is counted two times. Hence divide count by 2 and return.
Below is the implementation of above idea :
// C# implementation of simple method to
// find count of pairs with given sum
using System;
using System.Collections.Generic;
class GFG {
public static int[] arr = new int[] { 1, 5, 7, -1, 5 };
// Returns number of pairs in arr[0..n-1]
// with sum equal to 'sum'
public static int getPairsCount(int n, int sum)
{
Dictionary<int, int> hm
= new Dictionary<int, int>();
// Store counts of all elements
// in map hm
for (int i = 0; i < n; i++) {
// initializing value to 0,
// if key not found
if (!hm.ContainsKey(arr[i])) {
hm[arr[i]] = 0;
}
hm[arr[i]] = hm[arr[i]] + 1;
}
int twice_count = 0;
// iterate through each element and
// increment the count (Notice that
// every pair is counted twice)
for (int i = 0; i < n; i++) {
if (hm[sum - arr[i]] != 0) {
twice_count += hm[sum - arr[i]];
}
// if (arr[i], arr[i]) pair satisfies
// the condition, then we need to ensure
// that the count is decreased by one
// such that the (arr[i], arr[i])
// pair is not considered
if (sum - arr[i] == arr[i]) {
twice_count--;
}
}
// return the half of twice_count
return twice_count / 2;
}
// Driver Code
public static void Main(string[] args)
{
int sum = 6;
Console.WriteLine("Count of pairs is "
+ getPairsCount(arr.Length, sum));
}
}
// This code is contributed by Shrikant13
|
Count of pairs is 3
This article is contributed by Ashutosh Kumar.
修改錯誤code
[Theory]
[InlineData(new int[] { 6, 1, 3, 46 }, 47)]
// Returns number of pairs in arr[0..n-1]
// with sum equal to 'sum'
public static int getPairsCount2(int[] arr, int sum)
{
int n = arr.Length;
Dictionary<int, int> hm
= new Dictionary<int, int>();
// Store counts of all elements
// in map hm
for (int i = 0; i < n; i++)
{
// initializing value to 0,
// if key not found
if (!hm.ContainsKey(arr[i]))
{
hm[arr[i]] = 0;
}
hm[arr[i]] = hm[arr[i]] + 1;
}
int twice_count = 0;
// iterate through each element and
// increment the count (Notice that
// every pair is counted twice)
for (int i = 0; i < n; i++)
{
if (hm.ContainsKey(sum - arr[i]) && hm[sum - arr[i]] != 0)
{
twice_count += hm[sum - arr[i]];
}
// if (arr[i], arr[i]) pair satisfies
// the condition, then we need to ensure
// that the count is decreased by one
// such that the (arr[i], arr[i])
// pair is not considered
if (sum - arr[i] == arr[i])
{
twice_count--;
}
}
// return the half of twice_count
return twice_count / 2;
}
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